How to get raw pointer from cusp library matrix format

喜夏-厌秋 提交于 2019-12-31 06:22:24

问题


I need to get raw pointer from cusp library matrix format. For example:

cusp::coo_matrix<int,double,cusp::device_memory> A(3,3,4);

A.values[0] = 1;
A.row_indices[0] = 0;
A.column_indices[0]= 1;

A.values[1] = 2;
A.row_indices[1] = 1;
A.column_indices[1]= 0;

A.values[2] = 3;
A.row_indices[2] = 1;
A.column_indices[2]= 1;

A.values[3] = 4;
A.row_indices[3] = 2;
A.column_indices[3]= 2;

How can I get the raw pointer to row_indices, column_indices and values arrays? I need to pass them to my kernels and I'd like to avoid unnecesary data copying if possible.


回答1:


There are multiple ways to accomplish this. For example, if the you wish to start with the raw device data representation instead of the cusp data representation, you could use the methodology in the cusp views functionality.

If you have cusp data already, and you want to convert to raw data representation, we can use the fact that cusp is built on top of thrust. Here's a fully worked example:

$ cat t346.cu
#include <cusp/coo_matrix.h>
#include <cusp/print.h>

template <typename T>
__global__ void my_swap_kernel(T *a, T *b, unsigned size){
  int idx = threadIdx.x+blockDim.x*blockIdx.x;
  if (idx < size){
    T temp = b[idx];
    b[idx] = a[idx];
    a[idx] = temp;}
}



int main(void)
{
    // allocate storage for (4,3) matrix with 6 nonzeros
    cusp::coo_matrix<int,float,cusp::device_memory> A(4,3,6);

    // initialize matrix entries on host
    A.row_indices[0] = 0; A.column_indices[0] = 0; A.values[0] = 10;
    A.row_indices[1] = 0; A.column_indices[1] = 2; A.values[1] = 20;
    A.row_indices[2] = 2; A.column_indices[2] = 2; A.values[2] = 30;
    A.row_indices[3] = 3; A.column_indices[3] = 0; A.values[3] = 40;
    A.row_indices[4] = 3; A.column_indices[4] = 1; A.values[4] = 50;
    A.row_indices[5] = 3; A.column_indices[5] = 2; A.values[5] = 60;
    float *val0 = thrust::raw_pointer_cast(&A.values[0]);
    float *val3 = thrust::raw_pointer_cast(&A.values[3]);

    // A now represents the following matrix
    //    [10  0 20]
    //    [ 0  0  0]
    //    [ 0  0 30]
    //    [40 50 60]

    // print matrix entries
    cusp::print(A);
    my_swap_kernel<<<1,3>>>(val0, val3, 3);
    cusp::print(A);

    return 0;
}

$ nvcc -arch=sm_20 -o t346 t346.cu
$ cuda-memcheck ./t346
========= CUDA-MEMCHECK
sparse matrix <4, 3> with 6 entries
              0              0             10
              0              2             20
              2              2             30
              3              0             40
              3              1             50
              3              2             60
sparse matrix <4, 3> with 6 entries
              0              0             40
              0              2             50
              2              2             60
              3              0             10
              3              1             20
              3              2             30
========= ERROR SUMMARY: 0 errors
$


来源:https://stackoverflow.com/questions/22520669/how-to-get-raw-pointer-from-cusp-library-matrix-format

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